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but \(x^2+y^2\) will be equal to \((\frac{5}{3})^2 + (\frac{1}{3})^2 = \frac{25}{9}+\frac{1}{9} =\frac{26}{9}\) not even

Using the same logic, think what happens in stmnt 1 when \(x^2 = 5/3\) and \(y^2 = 1/3\).In this case, \(x^2 + y^2 = 2\) (even) but \(x^2 - y^2 = 4/3\) (not an even integer)but if \(x^2 = 12\) and \(y^2 = 6\), both \(x^2 + y^2\) and \(x^2 - y^2\) are even. So how can the answer be (A)?
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but \(x^2+y^2\) will be equal to \((\frac{5}{3})^2 + (\frac{1}{3})^2 = \frac{25}{9}+\frac{1}{9} =\frac{26}{9}\) not even

Using the same logic, think what happens in stmnt 1 when \(x^2 = 5/3\) and \(y^2 = 1/3\).In this case, \(x^2 + y^2 = 2\) (even) but \(x^2 - y^2 = 4/3\) (not an even integer)but if \(x^2 = 12\) and \(y^2 = 6\), both \(x^2 + y^2\) and \(x^2 - y^2\) are even. So how can the answer be (A)?

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06 Apr 2015, 23:11

My solution:For the first option: \(X^2 + Y^2 =\) even, then if you add or subtract \(2*X*Y\), you will get an even number aswell, so \(X^2 + Y^2 + 2*X*Y = (X + Y)^2\) is even and \((X - Y)^2\) is even too. Say both of them are \(2*k\) and \(2*b\) respectively.\(\sqrt{2*k}*\sqrt{2*b} = 2*\sqrt{k}*\sqrt{b}\) which is even, thus #1 is sufficient.For the second - well, its quite obvious: product of even number with whatever = even, thats true, thus sufficient

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06 Apr 2015, 23:15

Zhenek wrote:

My solution:For the first option: \(X^2 + Y^2 =\) even, then if you add or subtract \(2*X*Y\), you will get an even number aswell, so \(X^2 + Y^2 + 2*X*Y = (X + Y)^2\) is even and \((X - Y)^2\) is even too. Say both of them are \(2*k\) and \(2*b\) respectively.\(\sqrt{2*k}*\sqrt{2*b} = 2*\sqrt{k}*\sqrt{b}\) which is even, thus #1 is sufficient.For the second - well, its quite obvious: product of even number with whatever = even, thats true, thus sufficient

C it is then

Zhenek, this is look as you say that answer D is right: both statements are sufficient by themselves. Am I right?
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06 Apr 2015, 23:18

Harley1980 wrote:

Zhenek wrote:

My solution:For the first option: \(X^2 + Y^2 =\) even, then if you add or subtract \(2*X*Y\), you will get an even number aswell, so \(X^2 + Y^2 + 2*X*Y = (X + Y)^2\) is even and \((X - Y)^2\) is even too. Say both of them are \(2*k\) and \(2*b\) respectively.\(\sqrt{2*k}*\sqrt{2*b} = 2*\sqrt{k}*\sqrt{b}\) which is even, thus #1 is sufficient.For the second - well, its quite obvious: product of even number with whatever = even, thats true, thus sufficient

C it is then

Zhenek, this is look as you say that answer D is right: both statements are sufficient by themselves. Am I right?

Oh, yea, I guess I meant that indeed, not experienced with the gmat thing yet. Looks like my answer is wrong then, what a bummer. I guess I really need to pay attention to the given info (no information given about X and Y being non-integers, which completely blows my solution from the get-go: \(2*X*Y\) could be \(2*\sqrt{5}/2*\sqrt{3}/2\) which is not even remotely even )

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07 Apr 2015, 01:38

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Well then, lets amend my approach. Lets start from get-go.

Looking at 1st option and taking into account the fact that X and Y could be non-integers, we can't really say anything about the answer to the question with just option 1 on its own, I'd just call it insufficient right away after comming up with 2 different examples, one being integer and another - non-integer.Easiest ones that come into mind are: 1)X = 1, Y = 1: 2 is even, 0 is even2)X = \(\sqrt{5}/2\), Y = \(\sqrt{3}/2\): 2 is even, 1/2 is not evenThis makes #1 insufficient on its own.

#2 - same story, take fractions as second example and integers as first example, insufficient on its own.

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28 Nov 2017, 11:04

I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: \((X + Y)(X - Y)\) = Even?Which I rephrased too: \(X^2 + Y^2 - 2XY\) = Even?Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and \(2XY\) has to be Even. So I just went on to select D.

Please help me understand why I can't take this shortcut and I fell right into it.

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The question is: \((X + Y)(X - Y)\) = Even?Which I rephrased too: \(X^2 + Y^2 - 2XY\) = Even?Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and \(2XY\) has to be Even. So I just went on to select D.

Please help me understand why I can't take this shortcut and I fell right into it.

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I swear I have not been drinking. Just working too hard! Nonetheless, this is unacceptable of me. Thanks very much Karishma, and forgive my carelessness.

VeritasPrepKarishma wrote:

Hadrienlbb wrote:

I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: \((X + Y)(X - Y)\) = Even?Which I rephrased too: \(X^2 + Y^2 - 2XY\) = Even?Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and \(2XY\) has to be Even. So I just went on to select D.

Please help me understand why I can't take this shortcut and I fell right into it.

Show Tags

I swear I have not been drinking. Just working too hard! Nonetheless, this is unacceptable of me. Thanks very much Karishma, and forgive my carelessness.

VeritasPrepKarishma wrote:

Hadrienlbb wrote:

I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: \((X + Y)(X - Y)\) = Even?Which I rephrased too: \(X^2 + Y^2 - 2XY\) = Even?Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and \(2XY\) has to be Even. So I just went on to select D.

Please help me understand why I can't take this shortcut and I fell right into it.

Thanks,

Note that \((X + Y)*(X - Y) = X^2 - Y^2\)

It is better to make these errors during practice so that you do not make them during the actual exam!!
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